### Statistics

### Views

- Total Views
- 1,772
- Views on SlideShare
- 1,772
- Embed Views

### Actions

- Likes
- 0
- Downloads
- 89
- Comments
- 0

### Accessibility

### Categories

### Upload Details

Uploaded via SlideShare as Adobe PDF

### Usage Rights

© All Rights Reserved

Like this document? Why not share!

- Essential fluids by Keith Vaugh 17614 views
- Lec1 fluid mechanics by Keith Vaugh 815 views
- 0071487816 schaum s outline_of_fluid by ashoku2 4968 views
- Schaum's fluid mechanics - 260 by Chester Jed 3718 views
- Mecanica de fluidos potter wiggert ... by angelsantibanez 3055 views
- schaum's outline of fluid mechanics... by Jean Espinoza Vic... 5255 views
- Basic concepts of turbomachinery by Abinash Deuri 1233 views
- 1441976612 food processb by Gombiii 3426 views
- Fluid discharge by Keith Vaugh 17349 views
- Bernoulli\'s Principle by guestfda040 9551 views
- Bernoulli's Principle by eliseb 30951 views
- Fluid Mechanics by Mercedes Gontscharow 2668 views

**1,772**views

- Total Views
- 1,772
- Views on SlideShare
- 1,772
- Embed Views
- 0

- Likes
- 0
- Downloads
- 89
- Comments
- 0

No embeds

Uploaded via SlideShare as Adobe PDF

© All Rights Reserved

- FLUID MECHANICS {for energy conversion} Keith Vaugh BEng (AERO) MEng SME ASME
- OBJECTIVES Identify the unique vocabulary associated with ﬂuid mechanics with a focus on energy conservation } Explain the physical properties of ﬂuids and derive the conservation laws of mass and energy for an ideal ﬂuid (i.e. ignoring the viscous effects) Develop a comprehensive understanding of the effect of viscosity on the motion of a ﬂuid ﬂowing around an immersed body Determine the forces acting on an immersed body arising from the ﬂow of ﬂuid around itTo appreciate energy conversion such as hydro, wave, tidal and wind power a detailedknowledge of ﬂuid mechanics is essential.During the course of this lecture, a brief summary of the basic physical properties of ﬂuidsis provided and the conservation laws of mass and energy for an ideal (or inviscid) ﬂuid arederived.The application of the conservation laws to situations of practical interest are also exploredto illustrate how useful information about the ﬂow can be derived.Finally, the effect of viscosity on the motion of a ﬂuid around an immersed body (such as aturbine blade) and how the ﬂow determines the forces acting on the body of interest.
- BULK PROPERTIES OF FLUIDS Density (ρ) - Mass per unit volume ⎛ ⎜ ⎝ m⎞ ρ= ⎟ V⎠ } Pressure (P) - Force per unit area in a ﬂuid Viscosity - Force per unit area due to internal frictionDensity - Mass per unit volume of a ﬂuid. For the purposes of this module,Density is assumed to be constant i.e. that the ﬂow of the ﬂuid isincompressible (small variations in pressure arising from ﬂuid motion incomparison to atmospheric pressure).Pressure - Force per unit area, and acts in the normal direction to the surface ofa body in a ﬂuid. Its units are the pascal or N-m^2Viscosity - Force per unit area due to internal friction in a ﬂuid arising from therelative motion between neighbouring elements in a ﬂuid. Viscous forces act inthe tangential direction to the surface of a body immersed in a ﬂow. (consider adeck of cards.
- STREAMLINES & STREAM-TUBES }A useful concept to visualize a velocity ﬁeld is to imagine a set of streamlinespara;;e; to the direction of motion at all points in the ﬂuid. Any element of massin the ﬂuid ﬂows along a notational stream-tube bounded by neighbouringstreamlines. In practice, streamlines can be visualised by injecting smallparticles into the ﬂuid e.g. smoke can be used in wind tunnels to visualise ﬂuidﬂow around objects.
- MASS CONTINUITY Stream tube streamlines } velocity (u) Fluid stream ρuA = constAlso known as conservation of mass.Consider ﬂow along a stream tube in a steady velocity ﬁeld. The direction ofﬂow is parallel to the boundries and at any point within the stream tube, thespeed of the ﬂuid (u) and the cross-sectional area (A). Therefore the ﬂuid isconﬁned to the stream tube and the mass ﬂow per second is constant. Therefore:Density * Velocity * Cross-sectional Area = ConstantTherefore the speed of ﬂuid is inversely proportional to the cross-sectional areaof the stream tube.
- EXAMPLE 1An incompressible ideal ﬂuid ﬂows at a speed of 2 ms-1through a 1.2 m2 sectional area in which a constriction of }0.25 m2 sectional area has been inserted. What is the speedof the ﬂuid inside the constriction?Putting ρ1 = ρ2, we have u1A1 = u2A2 A1 ⎛ 1.2 ⎞ u2 = u1 = (2 m s )⎜ ⎟ = 9.6 m s A2 ⎝ 0.25 ⎠
- ENERGY OF A MOVING LIQUID The potential energy (PE) of liquid per unit mass ⎛ p⎞ } PE = g ⎜ z + ρg ⎟ p ⎝ ⎠ ρg and kinetic energy (KE) of liquid per unit mass u2 z KE = 2 total energy of liquid per unit mass ⎛ p u2 ⎞ Etotal = g⎜ z + + ⎝ ρ g 2g ⎟ ⎠Consider a liquid at a pressure p, moving with a velocity (u) at a height Z abovedatum levelIf a tube were inserted in the top of the pipe, the liquid would rise up the tube adistance of p/ρg and this is equivalent to an additional height of liquid relativeto datum level.Etotal equation represents the speciﬁc energy of the liquid. Each of the quantitiesin the brackets have the units of length and are termed heads, i.e Z is referred toas the potential head of the liquid, p/ρg as the pressure head and V2/2g as thevelocity head.
- ENERGY CONSERVATION IN AN IDEAL FLUID In many practical situations, viscous forces are much smaller that those due to gravity and pressure gradients over large regions of the ﬂow ﬁeld. We can ignore viscosity to a good approximation and } derive an equation for energy conservation in a ﬂuid known as Bernoulli’s equation. For steady ﬂow, Bernoulli’s equation is of the form; p 1 + gz + u 2 = const ρ 2 For a stationary ﬂuid, u = 0 p + gz = const ρFor a stationary ﬂuid, u = 0 everywhere in the ﬂuid therefore this equationreduces towhich is the equation for hydrostatic pressure. It shows that the ﬂuid at a givendepth is all the same at the same pressure.
- BERNOULLI’s EQUATION FOR STEADY FLOWConsider the steady ﬂow of an ideal ﬂuid in the control volume shown.The height (z), cross sectional area (A), speed (u) and pressure are denoted.The increase in gravitational potential energy of a mass δm of ﬂuid between z1and z2 is δmg(z2-z1)In a short time interval (δt) the mass of ﬂuid entering the control volume at P1is δm=ρu1A1δt and the mass exiting at P2 is δm=ρu2A2δt
- By energy conservation, this is equal to the increase in Potential Energy plus the increase in Kinetic Energy, therefore; p1 A1u1δ t + p2 A2u2δ t = Work done in pushing elemental mass δm δ mg ( z2 − z1 ) + δ m ( u2 − u12 ) 1 2 a small distance δs 2 δ s = uδ t Putting δ m = ρu1 A1δ t = ρu2 A2δ t and tidying δW1 = p1 A1δ s1 = p1 A1u1δ t p1 1 p 1 2 + gz1 + u12 = 2 + gz2 + u2 Similarly at P2 ρ 2 ρ 2 δW2 = p2 A2u2δ t Finally, since P1 and P2 are arbitrary The net work done is p 1 δW2 + δW2 = p1 A1u1δ t + p2 A2u2δ t + gz + u 2 = const ρ 2In order for the ﬂuid to enter the control volume it has to do work to overcomethe pressure p1 exerted by the ﬂuid. The work done in pushing the elementalmass δm a small distance δs = uδtat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity *Change in time. Remember F = P*ASimilarly, the work done in pushing the elemental mass out of the controlvolume at P2 is δW2=-p2A2u2δt.The net work done is δW1+δW2
- No losses between stations and u2 p1 u12 2 p2 u2 p2 z1 + + = z2 + + (1) ρg 2 ρg 2 u1 p1 Losses between stations and z2 z1 p1 u12 2 p2 u2 z1 + + = z2 + + +h (2) ρg 2 ρg 2 Energy additions or extractions p1 u12 win 2 p2 u2 wout z1 + + + = z2 + + + +h (3) ρg 2 g ρg 2 gConsider stations and of the inclined pipe illustrated. If there are nolosses between these two sections, and no energy changes resulting from heattransfer of work, then the total energy will remain constant, therefore stations and can be set equal to one another, equ (1)If however losses do occur between the two stations, then equation equ (2)When energy is added to the ﬂuid by a device such as a pump, or when energyis extracted by a turbine, these need also be accounted for equ (3). w is thespeciﬁc work in J/kgNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation
- QUESTION’s(a) The atmospheric pressure on a surface is 105 Nm-2. Assuming the water is stationary, what is the pressure at a depth of 10 m ? } (assume ρwater = 103 kgm-3 & g = 9.81 ms-2)(b) What is the signiﬁcance of Bernoulli’s equation?(c) Assuming the pressure of stationary air is 105 Nm-2, calculate the percentage change due to a wind of 20ms-1. (assume ρair ≈ 1.2 kgm-3)
- EXAMPLE 2 }A Pitot tube is a device for measuring the velocity in a ﬂuid.Essentially, it consists of two tubes, (a) and (b). Each tubehas one end open to the ﬂuid and one end connected to apressure gauge. Tube (a) has the open end facing the ﬂowand tube (b) has the open end normal to the ﬂow. In the casewhen the gauge (a) reads p = po+pU2 and gauge (b) readsp = po, derive an expression for the velocity of the ﬂuid interms of the difference in pressure between the gauges.
- Bernoulli’s equation p 1 + gz + u 2 = const ρ 2 p0 1 2 ps Applying bernoulli’s equation + U = ρ 2 ρ ⎡ 2 ( ps − p0 ) ⎤ 1 2 Rearranging U=⎢ ⎥ ⎣ ρ ⎦Consider the ﬂuid ﬂowing along the stream-line AB. In case (a), the ﬂuid slowsdown as it approaches the stagnation point B. Putting u = U, p = po at A and u= 0, p = ps at B and applying bernoulli’s equation we getIn example (and the previous question), ps is measured by tube (a) and p0measured by tube (b).NOTE - ps is larger than p0 by an amount ps - p0 = ρU2. The quantities ρU2,p0 and ps = p0 + ρU2 are called the dynamic pressure, static pressure and totalpressure respectively.
- VELOCITY DISTRIBUTION & FLOW RATE The velocity in a ﬂow stream varies considerably over the cross- section. A pitot tube only measures at one particular point, } therefore, in order to determine the velocity distribution over the entire cross section a number of readings are required.The velocity proﬁle is obtained by traversing the pitot tube along the line AA.The cross-section is divided into convenient areas, a1, a2, a3, etc... and thevelocity at the centre of each area is determined
- The ﬂow rate can be determine; V = a1u1 + a2u2 + a3u3 + ... = ∑ au If the total cross-sectional area if the section is A; the mean velocity, U = V = ∑ au A AThe velocity proﬁle for a circular pipe is the same across any diameter ΔV = ur × Δa where ur is the velocity at radius r V = ∑ ur × Δa and V= ∑u r × Δa πr2The velocity at the boundary of any duct or pipe is zero.
- EXAMPLE 3In a Venturi meter an ideal ﬂuid ﬂows with a volume ﬂowrate and a pressure p1 through a horizontal pipe of cross }sectional area A1. A constriction of cross-sectional area A2 isinserted in the pipe and the pressure is p2 inside theconstriction. Derive an expression for the volume ﬂow ratein terms of p1, p2, A1 and A2.
- From Bernoulli’s equation p1 1 2 p2 1 2 + u = + u ρ 2 1 ρ 2 2Also by mass continuity A1 u2 = u1 A2Eliminating u2 we obtain the volume ﬂow rate as ⎞ ⎡ 2 ( p1 − p2 ) ⎤ 1 2 ( V = A1u1 = ⎛ A1 A2 A12 − A ) 1 2 − 2 ⎝ 2 ⎠⎢ ρ ⎥ ⎣ ⎦
- Applying Bernoulli’s equation to stations and h p1 u12 2 p2 u2 z1 + + = z2 + + ρg 2 ρg 2 Given that meter is horizontal, u1 u2 this equation reduces to: p1 u12 p2 u2 2 + = + ρg 2 ρg 2 A1 Mass continuity A1u1 = A2u2 ⇒ u2 = u1 A2 2 p1 p2 u12 ⎛ A1 ⎞ u12 u12 ⎛ A12 − A2 ⎞ 2 ∴ − =h= − = ρg ρg 2g ⎜ A2 ⎟ ⎝ ⎠ 2g 2g ⎜ A2 ⎟ ⎝ 2 ⎠A venturi meter is a device which is used to measure the rate of ﬂow through apipe.NOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation
- A2 ∴u1 = ( 2gh ) ( A12 − A2 2 ) A1 A2V = A1u1 = ( 2gh ) = k h since A1 and A2 are constants (A 2 1 −A 2 2 )The pressure at the throat is slightly lower than the the theoretical due tofriction in the tapered pipe. As a consequence h becomes slightly largerand therefore resulting in volumetric ﬂow rate which is too high. Tocompensate for this, a coefﬁcient of discharge, cd is introduced. V = cd k h
- The pressure differential between the main and throat sections is generallymeasured with a mercury-and-water U-tube, therefore h = x ( S − 1)If a Venturi meter is inclined to the horizontal, i.e station has a height z1and station a height z2 then; V= k (h − ( z 2 − z1 ) )Show that the manometer reading for an inclined Venturi is the same as fora horizontal Venturi for a given ﬂow rate.
- ORIFICE PLATE Applying Bernoulli’s equation to stations and p1 u12 p2 u2 2 } + = + ρg 2 ρg 2 Mass continuity A1 A1u1 = A2u2 ⇒ u2 = u1 A2 ⎞ ⎡ 2 ( p1 − p2 ) ⎤ 1 2 ( ∴ V = A1u1 = ⎛ A1 A2 A12 − A ) 1 2 − 2 =k h ⎝ 2 ⎠⎢ ρ ⎥ ⎣ ⎦An Oriﬁce plate is another means of determining the ﬂuid ﬂow in a pipe andworks on the same principle of the Venturi meter.The position of the Vena contract can be difﬁcult establish and therefore thearea A2. The constant k is determined experimentally when it will incorporatethe coefﬁcient of discharge.If h is small so that ρ is approximately constant then the this equation can beused for compressible ﬂuid ﬂow.NOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation
- EXAMPLE 4In an engine test 0.04 kg/s of air ﬂows through a 50 mmdiameter pipe, into which is ﬁtted a 40 mm diameter Oriﬁce }plate. The density of air is 1.2 kg/m3 and the coefﬁcient ofdischarge for the oriﬁce is 0.63. The pressure drop acrossthe oriﬁce plate is measured by a U-tube manometer.Calculate the manometer reading.
- Mass ﬂow rate - m = ρV ⇒∴ V=0.033 m s 3 V= k h ⎛ π × 0.05 2 ⎞ ⎛ π × 0.04 2 ⎞ ⎜ ⎝ 4 ⎟⎜ ⎠⎝ 4 ⎟ ⎠ V= 1 2 × 9.81 × h ⎛ ⎛ π × 0.04 2 ⎞ 2 ⎛ π × 0.04 2 ⎞ 2 ⎞ 2 ⎜⎜ ⎝ 4 ⎟ −⎜ ⎠ ⎝ 4 ⎟ ⎟ ⎠ ⎠ ⎝ V = 0.007244 hThe actual discharge is 0.63 × 0.007244 h = 0.0333 ∴ h = 53.24m of airEquating pressures at Level XX in the U-tube 1.2g × 53.34 = 10 3 g × x ∴ x = 64 × 10 3 m
- DYNAMICS OF A VISCOUS FLUID The motion of a viscous ﬂuid is more complicated than that of an inviscid ﬂuid. } The viscous shear force per unit area in the ﬂuid is proportional to the velocity gradient y u ( y) = U F du u d = −µ =µ A dy d (0 ≤ y ≤ d) µ - coefﬁcient of dynamic viscosityLet us consider a simple case of laminar ﬂow between two parallel platesseparated by a small distance d.The upper plate moves at a constant velocity U while the lower plate remains atrest. At the plate ﬂuid interface in both cases there is no velocity due to thestrong forces of attraction. Therefore the velocity proﬁle in the ﬂuid is given by
- Typical ﬂow regimes Laminar ﬂow in a pipe Turbulent ﬂow in a pipe ρUL UL Reynolds Number, Re = = µ vA viscous ﬂuid can exhibit two different kinds of ﬂow regimes, Laminar andTurbulentIn laminar ﬂow, the ﬂuid slides along distinct stream-tubes and tends to be quitestable, Turbulent ﬂow is disorderly and unstableThe ﬂow that exists in an y given situation depends on the ratio of the inertialforce to the viscous force. The magnitude of this ratio is given by Reynoldsnumber, where U is the velocity, L is the length and v = µ/ρ is the kinematicviscosity. Reynolds observed that ﬂows at small Re where laminar while ﬂow athigh Re contained regions of turbulence
- Flow around a cylinder Flow around a cylinder for a for an inviscid ﬂuid viscous ﬂuidFor the inviscid ﬂow the velocity ﬁelds in the upstream and downstreamregions are symmetrical, therefore the corresponding pressure distribution issymmetrical. It follows that the net force exerted by the ﬂuid on the cylinder iszero. This is an example of d’Alembert’s paradoxFor a body immersed in a viscious ﬂuid, the component of velocity tangential tothe surface of the ﬂuid is zero at all points. At large Re numbers (Re>>1), theviscous force is negligible in the bulk of the ﬂuid but us very signiﬁcant in aviscous boundary layer close to the surface of the body. Rotational componentsof ﬂow know as vorticity are generated within the boundary layer. As a certainpoint, the seperation point, the boundary layer becomes detached from thesurface and vorticity are discharged into the body of the ﬂuid. The vorticity aretransported downstream side of the cylinder in the wake. Therefore the pressuredistributions on the upstream and downstream sides of the cylinder are notsymmetrical in the case of a viscous ﬂuid. As a result, the cylinder experiencesa net force in the direction of motion known as drag. In the case of a spinningcylinder, a force called life arises at right angles to the direction of ﬂow.
- Lift and Drag Lift = 1 2 C L ρU 2 A Drag = 1 2 C D ρU 2 ACL and CD are dimensionless constants know as the coefﬁcient of Lift and thecoefﬁcient of Drag respectively.Lift and drag can be changed by altering the shape of a wing i.e. ailerons. Forsmall angles of attack, the pressure distribution ont he upper surface of anaerofoil is signiﬁcantly lower than that on the lower surface which results is in anet lift force.
- Variation of the lift and drag coefﬁcients CL and CD with angle of attackCL and CD are dimensionless constants know as the coefﬁcient of Lift and thecoefﬁcient of Drag respectively.Lift and drag can be changed by altering the shape of a wing i.e. ailerons. Forsmall angles of attack, the pressure distribution on the upper surface of anaerofoil is signiﬁcantly lower than that on the lower surface which results is in anet lift force.
- Bulk properties of ﬂuidsStreamlines & stream-tubesMass continuityEnergy conservation in an ideal ﬂuidBernoulli’s equation for steady ﬂow Applied example’s (pitot tube, Venturi meter & questionsDynamics of a viscous ﬂuid Flow regimes Laminar and turbulent ﬂow Vortices Lift & Drag
- Andrews, J., Jelley, N., (2007) Energy science: principles, technologies andimpacts, Oxford University PressBacon, D., Stephens, R. (1990) Mechanical Technology, second edition,Butterworth HeinemannBoyle, G. (2004) Renewable Energy: Power for a sustainable future, secondedition, Oxford University PressÇengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal ﬂuidsciences, Third edition, McGraw HillTurns, S. (2006) Thermal ﬂuid sciences: An integrated approach, CambridgeUniversity PressTwidell, J. and Weir, T. (2006) Renewable energy resources, second edition,Oxon: Taylor and FrancisIllustrations taken from Energy science: principles, technologies and impacts & Fundamentals of thermal ﬂuid science

Full NameComment goes here.